Post by: HarveyD on September 09, 2011, 08:14:55 pm
a and b are just symbols
2) A parcel rests on a stationary conveyor belt that slopes at an inclination of 30 degrees to the horizontal. The belt begins to move upwards with an increasing acceleration, and when this reaches g/4, the parcel begins to slip.
Calculate the co-efficient of friction between the parcel and the belt.
Post by: b^3 on September 09, 2011, 08:54:17 pm
Refer to image 1 below.
Friction will be up the plane since it needs to balance the force of gravity down the plane, the reason it needs to balance it is because it is in equlibrium.
We need to find r to find μ, so to do that we need to resolve the forces perpendicular to the plane.
R=mgcos(a) ....1
Next resolve the force parallel to the plane
μR-mgsin(a)=0
so μR=mgsin(a) ....2
or μ=(mgsin(a))/R
Now substitute 1 into 2
μ=(mgsin(a))/mgcos(a)
so the mg's cancel and sin on cos is tan
so μ=tan(a)
Now finding acceleration.
So inclination is increased to b and the object is no longer in equilibrium. So we need to resolee the forces but this time letting the parallel force equal ma (i.e. Fnet)
Perpendicular
so R=mgcos(b)
Parallel
take mgsin(b) first as it will be large as the acceleration is down the plane
mgsin(b)-μR=ma
mgsin(b)-μmgcos(b)=ma
so a=mg(sin(b)-μcos(b))/m
so then a=g(sin(b)-μcos(b)) m/s2
***Alternate solution Essential uses****
Yeh they did it a silly way. this is what they did from the mgsin(b)-μ*R=ma part
then
then use the sin(x-y)=sin(x)cos(y)-cos(x)sin(y)
to get a=g*sin(b-a)/cos(a)
2) We know that when the acceleration reachers g/4, the particle slips, in other words the maximum friction force has been exerted. So the Frmax=μR=the force under an acceleration of g/4
=mg/4
Now we need just μ so we need to find R and sub it in.
R will be mgcos(30)=mgroot(3)/2 N
Resolve parallel to surface
μR-mgsin(30)=mg/4
μroot(3)/2mg-mg/2=mg/4
cancel the mg's
root(3)μ/2=3/4
μ=(3/2root(3))*root(3)/root(3) (rationalising)
μ=root(3)/2
Post by: HarveyD on September 09, 2011, 09:02:40 pm
except answers had
acceleration = (g/cos a) sin (b - a)
:S
Post by: b^3 on September 09, 2011, 09:04:44 pm
Post by: HarveyD on September 09, 2011, 09:12:02 pm
Post by: b^3 on September 09, 2011, 09:12:36 pm
then
then use the sin(x-y)=sin(x)cos(y)-cos(x)sin(y)
to get a=g*sin(b-a)/cos(a)
There, was the second one right?
Post by: b^3 on September 09, 2011, 09:17:19 pm
Post by: b^3 on September 09, 2011, 09:20:51 pm
Post by: HarveyD on September 09, 2011, 09:27:43 pm
i thought it would be mg/4 = uR + mg/2, since the belt is going up the incline and friction acts down?
probably misinterpreted it though :/
Post by: b^3 on September 09, 2011, 09:32:03 pm
EDIT: Keep in mind the arrows for my vectors are not to scale.
Post by: HarveyD on September 09, 2011, 09:38:50 pm
thanks so much!
Post by: HarveyD on September 09, 2011, 11:14:00 pm
A train that is moving with uniform acceleration is observed to take 20s and 30s to travel successive half kilometres. How much farther will it travel before coming to rest if the acceleration remains constant?
(Question 13 of Chapter Review)
How would I do this :S
Post by: Mao on September 10, 2011, 04:40:12 am
Given that {t=20, s=500} and {t=50, s=1000}, you can use the above to find u and a. The rest is trivial.
Post by: HarveyD on September 15, 2011, 04:50:04 pm
A 500g ball hits a lake with a speed of 8m/s water resistance is kt Newtons. The ball reaches a maximum depth of 80cm.
Find k and time taken to reach this depth.
Post by: b^3 on September 15, 2011, 04:55:38 pm
so F=mg-kt
F=0.5g-kt
F=ma
a=F/m
a=(0.5g-kt)/0.5
a=g-2kt
Intergrate this to get velocity
v=gt-kt2+d
t=0, v=8
8=0+0+d
d=8
so v=gt-kt2+8
Intergrate that to get displacement
s=gt2/2 -kt3/3+8t+c
s=0,t=0
0=0-0+0+c
c=0
v=0,s=0.8
find t when s=0.8 (i.e. v=0)
substitute that value of t in to find k
I hope that is all right, I feel like I have made a mistake somewhere.
I have done something wrong somewhere, give me a min.
Post by: luken93 on September 15, 2011, 04:59:53 pm
a = kt/0.5 = 2kt
x = t/2(u +v)
0.8 = t/2(8 + 0)
4/5 = 4t
t = 1/5 sec
x = ut + 1/2at^2
4/5 = 8/5 + 1/2 * 2k/5 * 1/25
4/5 = 8/5 + 2k/250
200 = 400 + 2k
k = -100
Hopefully I haven't made any mistakes :)
Woops, not constant acceleration :S
Post by: b^3 on September 15, 2011, 05:05:58 pm
Post by: HarveyD on September 15, 2011, 05:30:45 pm
this was what i did
Resultant = 0.5g - kt
m = 0.5 kg
so accel = g - 2kt
vel = gt - kt^2 + 8
v = 0 when x = 80
Setting up simultaneous equations:
0 = gt - kt^2 + 8
80 = g/2kt^2 - 1/3kt^3 + 8t
therefore t = 5.55
and k = 2.02
is that right? :S
Post by: b^3 on September 15, 2011, 05:32:45 pm
anyway, Yes except it would be 0.8 m for thw displacement since the velocity is in m/s not cm/s. i.e.
k=0.603 and t=17.023
or k=2.585 and t=-0.691
reject the bottom one since t>0
Post by: HarveyD on September 15, 2011, 05:35:23 pm
ah yes, thanks!
Post by: HarveyD on September 15, 2011, 06:58:51 pm
Does that mean we take "projected" as no force being applied....
Post by: luken93 on September 15, 2011, 07:20:47 pm
In the Essentials Textbook, they use the term "projected" up an incline, but in the solutions they just use the force acting down the incline i.e. mgsin(theta) to work out the acceleration.Yep, if a force is applied initially, it is never in the force diagram. Crops up every so often in exams as well to trick people.
Does that mean we take "projected" as no force being applied....
Post by: b^3 on September 15, 2011, 07:22:16 pm
EDIT: beaten by 48 seconds.
Post by: HarveyD on September 15, 2011, 10:03:34 pm
Is the solution at the back correct, cause i cant seem to get it ><
i keep getting 5/4 rather than 5/4 x g
using this:
u = Answer for b
v = 0
a = -0.2 x g
s = ?
v^2 = u^2 + 2as
so the g's would cancel?
Post by: b^3 on September 15, 2011, 10:10:03 pm
Post by: HarveyD on September 15, 2011, 11:01:55 pm
Find when it comes to rest:
Teacher did this:
0 = 40 (1 - 2e^-0.2t)
but shouldnt it be
800a = 2400(1-2e^-0.2t)
then anti diff
then solve for when v = 0?
Post by: xZero on September 15, 2011, 11:34:57 pm